Find a homogeneous linear differential equation with constant coefficients whose general solution is given by y=c1e−xcosx+c2e−xsinx.

Question

Find a homogeneous linear differential equation with constant coefficients whose general solution is given by
y=c1e−xcosx+c2e−xsinx.

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Tài Đức 1 year 2021-09-04T15:15:49+00:00 1 Answers 16 views 0

Answers ( )

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    2021-09-04T15:17:03+00:00

    Answer:

    [tex]y” + 2y’ + 2y = 0[/tex]

    Step-by-step explanation:

    Given

    [tex]y=c_1e^{-x}cosx+c_2e^{-x}sinx[/tex]

    Required

    Determine a homogeneous linear differential equation

    Rewrite the expression as:

    [tex]y=c_1e^{\alpha x}cos(\beta x)+c_2e^{\alpha x}sin(\beta x)[/tex]

    Where

    [tex]\alpha = -1[/tex] and [tex]\beta = 1[/tex]

    For a homogeneous linear differential equation, the repeated value m is given as:

    [tex]m = \alpha \± \beta i[/tex]

    Substitute values for [tex]\alpha[/tex] and [tex]\beta[/tex]

    [tex]m = -1 \± 1*i[/tex]

    [tex]m = -1 \± i[/tex]

    Add 1 to both sides

    [tex]m +1= 1 -1 \± i[/tex]

    [tex]m +1= \± i[/tex]

    Square both sides

    [tex](m +1)^2= (\± i)^2[/tex]

    [tex]m^2 + m + m + 1 = i^2[/tex]

    [tex]m^2 + 2m + 1 = i^2[/tex]

    In complex numbers:

    [tex]i^2 = -1[/tex]

    So, the expression becomes:

    [tex]m^2 + 2m + 1 = -1[/tex]

    Add 1 to both sides

    [tex]m^2 + 2m + 1 +1= -1+1[/tex]

    [tex]m^2 + 2m + 2= 0[/tex]

    This corresponds to the homogeneous linear differential equation

    [tex]y” + 2y’ + 2y = 0[/tex]

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